All categories

2 years ago

14/30÷14.00 show all work

Mathematics

1 answer:

kati45 [8]2 years ago

5 0

1 Simplify \frac{14}{30}

30

14

 to \frac{7}{15}

15

 .

\frac{7}{15}\div 14.00

15

 ÷14.00

2 Use this rule: a\div \frac{b}{c}=a\times \frac{c}{b}a÷

 =a×

 .

\frac{7}{15}\times \frac{1}{14.00}

15

 ×

14.00

3 Use this rule: \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}

 ×

 =

bd

ac

 .

\frac{7\times 1}{15\times 14.00}

15×14.00

7×1

4 Simplify 7\times 17×1 to 77.

\frac{7}{15\times 14.00}

15×14.00

5 Simplify 15\times 14.0015×14.00 to 210210.

\frac{7}{210}

210

6 Simplify.

1/30

You might be interested in

Express the following complex number in trigonometric form: 3 - 3i and find the 4th roots.

Sonja [21]

Answer:

$z=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right)$

$z_1=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{7\pi}{16}+i\sin\dfrac{7\pi}{16}\right).$

$z_2=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{15\pi}{16}+i\sin\dfrac{15\pi}{16}\right).$

$z_3=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{23\pi}{16}+i\sin\dfrac{23\pi}{16}\right).$

$z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).$

Step-by-step explanation:

The complex number $z=3-3i$ has the real part $Re\ z=3$ and the imaginary part $Im\ z=-3.$

Hence,

$|z|=\sqrt{(Re\ z)^2+(Im\ z)^2}=\sqrt{3^2+(-3)^2}=\sqrt{9+9}=3\sqrt{2},\\ \\\cos \varphi=\dfrac{Re\ z}{|z|}=\dfrac{3}{3\sqrt{2}}=\dfrac{\sqrt{2}}{2},\\ \\\sin \varphi=\dfrac{Im\ z}{|z|}=\dfrac{-3}{3\sqrt{2}}=-\dfrac{\sqrt{2}}{2}.$

From the last two equalities, $\varphi =\dfrac{7\pi}{4}$ and the trigonometric form is

$z=|z|(\cos\varphi+i\sin\varphi)=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right).$

The square roots can be calculated using the formula:

$\sqrt[4]{z}=\left\{\sqrt[4]{|z|}\left(\cos\dfrac{\varphi+2\pi k}{4}+i\sin\dfrac{\varphi+2\pi k}{4}\right),\text{ where }k=0,1,2,3\right\}.$

At k=0:

$z_1=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}}{4}+i\sin\dfrac{\frac{7\pi}{4}}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{7\pi}{16}+i\sin\dfrac{7\pi}{16}\right).$

At k=1:

$z_2=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+2\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+2\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{15\pi}{16}+i\sin\dfrac{15\pi}{16}\right).$

At k=2:

$z_3=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+4\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+4\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{23\pi}{16}+i\sin\dfrac{23\pi}{16}\right).$

At k=3:

$z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+6\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+6\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).$

3 0

3 years ago

50 points! Mhanifa please help! I will mark brainliest! Random answers will be reported guys so don’t answer if you don’t know i

nlexa [21]

Answer:

136

Step-by-step explanation:

For the second question, it is 136 because for triangles all side when added always equal to 180, so if you add 24 and 20 (which = 44) and subtract it from 180, it gives you 136, which is the answer and the angle of n.

8 0

2 years ago

Read 2 more answers

Will make brainleist find the value of ar in the figure below if re is parallel to bc show your work

Rainbow [258]

Answer:

AR/AB = AE/AC

Substitute and group the like terms

you will find that x=5

Step-by-step explanation:

therefore, AR = x-3

. AR = 5-3

AR = 2

3 0

2 years ago

Help me with this piecewise function

Thepotemich [5.8K]

Answer:

$\displaystyle f(x)=\left \{ \begin{array}{ll}{{-(x+3)^2+3&\text{for $x$

Step-by-step explanation:

The left section of the function is a parabola with vertex (-3, 3) and a vertical scale factor of -1. You can tell that -1 is the scale factor because the parabola opens downward and 1 horizontal unit from the vertex, the function is 1 unit vertically different from the vertex. (The vertical difference is the scale factor.)

The right section is an absolute value function with a vertex at (0, -1) and a scale factor of 1/2. You can tell the scale factor is 1/2 because the rise/run of the lines is 1/2.

The left section is defined for x < -2; the right section is defined for x ≥ -2.

The vertex locations show the function translation, which is applied in the usual way:

f(x) translated to (h, k) is <em>f(x -h) +k</em>.

$\displaystyle f(x)=\left \{ \begin{array}{ll}{{-(x+3)^2+3&\text{for $x$